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B E L A R U S

M i n i s t r y o f E d u c a t i o n

Chemistry Olympiad

National Final

Theoretical Examination

Gomel, March 29 - April 03, 2010

44th Belarus Chemistry Olympiad, National Final (Gomel, March 29–April 3, 2010)

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BELARUSIAN CHEMISTRY OLYMPIAD

Belarusian Chemistry Olympiad is an annual science competition held continuously since the early 1970s. The olympiads are organized and coordinated by faculty members of chemistry departments of major Belarusian universities with financial support from the Belarus Ministry of Education. Most participants are high school students of grades 9 through 11. It is not unusual, however, that exceptionally gifted ninth-grade students com-pete in the National Final.

Selection of nominees for the National Final within each grade is conducted in two rounds. In December-January, competitions within local school districts determine qualified participants for regional olympiads. The regional olympiads are held in February in each of the six provinces associated with the country's principal cities. The capital city of Minsk holds a separate competition elevated in status to regional. Approximately 30 students from each grade are nominated to sit for the National Final.

The National Final consists of a theoretical exam and a laboratory practical. The five-hour theoretical exam includes a 10-item multiple choice test and a set of 5 problems. The laboratory practical involves 1-2 experimental problems and lasts between 4 and 5 hours. The winners of the National Final attend a study camp at the Belarusian State Uni-versity in Minsk where they undergo additional training in preparation for the International Chemistry Olympiad.

44rd Belarus Chemistry Olympiad, National Final (Gomel, March 29–April 3, 2010)

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GRADE 9

Problem 9-1 The freezing point of a solution is defined as the temperature at which first crystals of the solvent are formed. The freezing point of a pure solvent is equal to its melting point. It is well known that the freezing point of a solution is depressed relative to the freezing point of the pure solvent. The freezing point depression, ΔT, is related to the molality of the solu-tion, b, by the equation

ΔT = Kb, (equation 1) where K is the cryoscopic constant.

1) When a pure solvent freezes, how does its temperature change in the process? 2) When a solution freezes, how does its temperature change in the process? 3) The freezing point of a certain aqueous solution of NaCl is –0.500°C. Calculate

the mass percent of NaCl in this solution. 4) The cryoscopic constant is related to the enthalpy of fusion of the solvent (ΔmH)

by the equation K = RTm2M/ΔmH, where Tm is the melting point of the solvent and

M is its molar mass. For water, K = 1.86 K·kg·mol–1. Calculate the molar enthalpy of fusion of water.

5) A more accurate relationship between the freezing point depression and the con-centration is ΔT = RTm

2xs/ΔmH, (equation 2) where xs is the mole fraction of the solute in the solution. Show that for small xs, equation 2 reduces to equation 1.

6) The enthalpy of fusion of a certain substance is 10.14 kJ/mol. A sample of this substance contains a small amount of an impurity. When heated at 181.85 K, 28% of the sample is in the liquid phase; at 182.25 K, this fraction increases to 53.0%. Calculate the melting point of the pure substance, the mole fraction of the impurity in the sample, and the melting point of the sample.

Problem 9-2 Three different minerals contain the same mole percent of lead. One of these minerals was first characterized by the German chemist Martin Klaproth in 1784, and it was named pyromorphite by the German mineralogist Johann Hausmann in 1813. Pyromorphite is 2.614% chlorine and 6.851% phosphorus by mass. It often occurs naturally with the sec-ond mineral called mimetite. This name derives from the Greek word mimethes (μιμητήζ), meaning “imitator”. The mass percent of chlorine in mimetite is 2.382%. The third mineral was originally discovered in 1801 in Mexico by the Spanish mineralogist Andrés Manuel del Río. He called the mineral “brown lead” and pointed out that it contained a new ele-ment, which he named erythronium. Today, this mineral is the principal source of “erythronium”. The mass percent of chlorine in “brown lead” is 2.503%.

1) Deduce the chemical formulas of the three minerals in question. 2) Write balanced equations for the reactions of pyromorphite with excess amounts

of dilute and concentrated hydrochloric acid. 3) What is the theoretical mass of “erythronium” that can be obtained from 10 metric

tons of ore containing “brown lead”, given that the ore is 12.0% barren matter? Assume that the process involves four steps, each with a practical yield of 89%.

Problem 9-3 The enthalpy change for a chemical reaction is defined as the sum of the enthalpies of for-mation of the products minus the sum of the enthalpies of formation of the reactants, each


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multiplied by the corresponding stoichiometric coefficient. Enthalpies of formation of ele-mentary substances in their standard states are taken to be zero. Each of three unknown hydrocarbons contains two carbon atoms per molecule. The standard enthalpies of formation (ΔfH0) of these hydrocarbons, arranged in the order of in-creasing number of hydrogen atoms, are 227, 52, and –85 kJ/mol, respectively. From standard tables of thermodynamical properties of compounds, ΔfH0(CO2(g)) = –395 kJ/mol, ΔfH0(H2O(g)) = –242 kJ/mol.

1) Write the structural formulas of the three hydrocarbons in question. Name these compounds. For each compound, suggest a method of preparation using com-mon reagents available in the chemical laboratory.

2) Write balanced equations to describe combustion of each of these hydrocarbons. Calculate the standard enthalpy change for each reaction.

3) For a hydrocarbon of the general formula CnH2n, what is the mole ratio of the hy-drocarbon and oxygen gas maximizing the heat of combustion?

4) Each of the above three hydrocarbons is combined with a stoichiometric amount of oxygen gas. Calculate the ratio of the standard enthalpies of combustion of equal volumes of these stoichiometric mixtures.

5) Which of these three stoichiometric mixtures produces the hottest flame? Give your reasoning.

Problem 9-4 Compound A is a hydroxide of an unknown metal. When A is heated in an inert atmos-phere, a solid residue (compound B) and a mixture of gases are formed. Compound B is 27.6% oxygen by mass. The gaseous mixture has a density of 4.20 × 10–4 g/cm3 at 400 K and 110 kPa.

1) Determine the qualitative and quantitative composition of the gaseous mixture (in terms of mass and mole fractions).

2) Deduce the chemical composition of compounds A and B. Write a balanced equation to represent the reaction described in the problem statement.

3) What is the shortest synthetic route to the hydroxide A starting from the corre-sponding metal? Write a balanced equation for each step.

Problem 9-5 Calcium ions are neither reducing nor oxidizing agents in aqueous solution, so they cannot be detected by redox titration. However, they can be determined indirectly by substitution titration.

A 14.64 g sample of limestone was dissolved in hydrochloric acid. The solution was filtered, transferred quantitatively to a 100 mL graduated flask, and diluted to the mark with distilled water. A 25.00 mL sample of the prepared solution was neutralized with aqueous ammonia (in the presence of methyl orange) and combined with 50.0 mL of a 2.000 M aqueous solution of sodium oxalate, Na2C2O4 (the sodium salt of oxalic acid). The result-ing white precipitate was separated and rinsed on filter paper. The filtrate was combined with the rinse solution, transferred to a 100 mL graduated flask, and diluted to the mark with distilled water. A 10.00 mL sample of this solution was titrated in the presence of sul-furic acid with 14.85 mL of a 0.1844 M aqueous solution of potassium permanganate.

1) Write balanced equations to describe the reactions that occur in this analysis, as-suming that oxalic acid is oxidized by KMnO4 to carbon dioxide and water.

2) Calculate the mass percent of CaCO3 in the initial sample of limestone.


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GRADE 10

Problem 10-1 An unknown aromatic hydrocarbon Х is 89.49% carbon by mass. It is known that the vapor density of X at 1 atm and 25°C does not exceed 10 g/L.

1) Determine the molecular formula of Х. 2) Draw all possible structural formulas of Х, given that free-radical chlorination of X

yields only one monochlorosubstituted derivative. 3) Identify compound X, given that it is not readily oxidized with aqueous KMnO4. 4) Compound X can be nitrated with a nitrating mixture, and the product of that reac-

tion can be oxidized with hot aqueous HNO3. These two reactions can also be also carried out in the reverse order. What are the final products in each case? Write the reaction scheme for each case.

5) Are there any optically active arenes isomeric with compound X? If so, write their stereochemical formulas showing the absolute configuration of each isomer. If not, explain why.

Problem 10-2 A certain aqueous solution contains three oxoacids of phosphorus: X, Y, and Z. Acid X is 4.58% hydrogen and 46.9% phosphorus by mass; acid Y is 37.8% phosphorus by mass; the total number of atoms in a molecule of Z is 14.3% greater than in a molecule of Y.

6) Determine the molecular formulas of X, Y, and Z. Show your calculations. 7) Write the structural formulas of X, Y, and Z and arrange these acids in the order

of increasing values of their first dissociation constants. Give your reasoning. The complete neutralization of a 50.0 cm3 sample of the initial solution requires 9.35 cm3 of a 10.0% aqueous solution of NaOH (with a density of 1.07 g/cm3). Two other 50.0 cm3 samples were subjected to the following procedures and then titrated with 0.100 M sodium thiosulfate:

Sample Procedure V(Na2S2O3), cm3

1 The sample is treated with 200 mL of a 0.100 М solution of I2, and the mixture is heated for 24 hours 80.00

2 The sample is neutralized with NaOH, the resulting mix-ture is treated with 200 mL of a 0.100 M solution of I2, and then is allowed to stand for 4 hours

360.0

8) Write balanced equations to describe the reactions that occur in this analysis. 9) What are the molar concentrations of X, Y, and Z in the initial solution? 10) How would you prepare acid X from phosphorus in the chemical laboratory?

Problem 10-3 A mixture of nitrogen and hydrogen gases with a total volume of 0.500 dm3 (measured at S.T.P.) was passed over a heated platinum catalyst. One half of the gaseous mixture of products was bubbled through 10 cm3 of a 4.39% aqueous solution of HCl with a density of 1.020 g/cm3. Titration of the excess acid required 18.30 cm3 of 0.5486 M NaOH. The second half of the mixture was burned in an excess of oxygen gas and the combustion products were passed through a column filled with P2O5. As a result, the mass of the col-umn increased by 0.08438 g.

1) Write balanced equations to represent the reactions that occur in this experiment. 2) Calculate the mass percent of HCl in the solution before the titration. 3) Calculate the mass percent of nitrogen gas in the initial mixture.

Problem 10-4 Phosphorus is an undesirable component of metal alloys because it makes them brittle at


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low temperatures. In iron-based alloys, phosphorus normally occurs in the form of iron phosphides or as a solid-state solution. High phosphorus concentrations in Fe-P alloys can be determined by the following method. A sample of cast iron is dissolved in an excess of hot nitric acid. The resulting solution is replenished with HNO3, and then an excess of aqueous ammonium molybdate is added to the reaction mixture. This causes all of the phosphorus to precipitate in the form of a bright-yellow heteropoly acid (NH4)3H4[P(Mo2O7)6], which is insoluble in nitric acid. The precipitate is separated, rinsed on filter paper, and dissolved in an excess of aqueous ammonia. The resulting solution, which contains ammonium molybdate and ammonium hydrophosphate, is treated with aqueous solutions of magnesium chloride and ammonia. This produces a precipitate of magnesium-ammonium orthophosphate hexahydrate, which is rinsed, filtered off, and heated to constant mass at a high temperature. The mass loss constitutes 54.66%. The resulting solid residue is allowed to cool and then is weighed. It should be noted that if the oxidizing reagent used in the first step of the analysis is not suf-ficiently strong, a substantial fraction of the phosphorus may be lost, which will distort the final result of the analysis.

1) Assuming that all of the phosphorus in the sample is in the form of Fe3P, write balanced equations to represent the reactions that occur in this analysis.

2) Explain why the use of a weaker oxidizing reagent in the first step can affect the result of the determination. Write a chemical equation to represent the corre-sponding reaction.

3) Calculate the mass percent of phosphorus in a 14.66 g sample of cast iron, given that the mass of the solid residue in the final step is 179.6 mg.

Problem 10-5 A certain crystalline compound X can be prepared by fusing calcium metal with boron at 900°C. Compound X has a density of 2.44 g/dm3 and finds many industrial applications. For example, it is used as a deoxidation agent in the production of oxygen-free copper. The unit cell of X is a cube with a side of 415 pm. The corner of each cube is occupied by a calcium atom while a group of boron atoms occupies the center.

1) Determine the molecular formula of X and name this compound. Show your cal-culations.

2) Write the chemical formulas of the two substances into which compound X may transform during the production of “oxygen-free copper”.

3) Suggest a different method for synthesizing compound X in the chemical labora-tory. Write chemical equations for the reactions involved in this synthesis and in-dicate the conditions under which they take place.

4) What is the spatial arrangement of the boron atoms occupying the center of the unit cell?

GRADE 11

Problem 11-1

“There is no disputing about tastes, and therefore there is always bragging, brawling, and rioting about tastes.” G. K. Chesterton

A white crystalline compound X draws attentions of scientists working in various fields. Some people find that X tastes very bitter even in minute concentrations, while oth-ers find it virtually tasteless. Yet another group asserts that X tastes sweet. It is believed that the ability to perceive X determines whether an individual can become a professional taster. Studies have shown that the ability to taste X depends on the genetic makeup of


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the person, and so it is hereditary. Compound X can be synthesized according to the following scheme. The starting

material is a certain hydrocarbon A which is 92.3% carbon by mass and has a vapor den-sity not exceeding 4 g/L.

A B C D E XHNO3

H2SO4

Fe/HCl 1) NaOH

2) CS2, t0

HCl, t0 NH3

- C.

1) Deduce the structural formulas of compounds А – Е and Х, given that D is 14.05% sulfur and 12.27% nitrogen by mass, and that compound X is 55.24% carbon by mass.

2) Name the compound Х. 3) What is the theoretical mass of X that can be obtained from 1.00 kg of A? 4) Compound X can be also prepared by heating C with a certain potassium salt

(compound Y), which is 12.36% carbon by mass. Identify compound Y and write a balanced equation for this reaction. What historically significant synthesis is analogous to this method of preparation of X?

Problem 11-2 An unknown binary compound A is a solid at S.T.P and contains more than 10% hydrogen by mass. A strong reducing agent, compound A reacts with water to give an elementary substance B. When A is heated under pressure in an atmosphere of CO2, the only product is a colorless crystalline compound C, which is 61.6% oxygen by mass. When C reacts with dilute sulfuric acid, the product is an organic compound D. However, when C reacts with concentrated H2SO4, the product is a gas E which has a lower density than air.

1) Identify the lettered compounds and give their names. 2) Write chemical equations to describe the reactions that occur in this experiment. 3) Write an equation for the reaction of AlCl3 with an excess of A. What is the practi-

cal use of the product of this reaction? 4) Assuming that each of the two elements comprising compound A occurs naturally

in the form of two isotopes with relative isotopic masses differing by 1, calculate the relative abundances of the four possible isotopomers of A.

Problem 11-3 When organizing safe storage and handling of combustible solid materials, it is important to take into account their calorific values. These values are routinely included in material safety data sheets. The questions below refer to the following table.

Substance Calorific value Q, MJ/kg Polyethylene 47.14 Natural rubber 44.73

1) What is the definition of the calorific value? 2) Using the data provided in the table, estimate the calorific value of butadiene-

based synthetic rubber. Indicate all the assumption you have made. 3) Arrange the following substances in the order of increasing calorific value: cellu-

lose, polymethyl methacrylate, natural rubber (polyisoprene), and polystyrene. Give your reasoning.

A common model of fire extinguishers consists of two chambers. The main chamber con-tains 8 kg of a 5% aqueous solution of sodium bicarbonate and a small amount of a sur-factant additive, while the auxiliary chamber contains 400 g of aqueous sulfuric acid. When the fire extinguisher is activated, the acid flows into the main chamber. The evolving gas causes the mixture to stir vigorously. Due to a pressure build-up, the foam is expelled from the main chamber and can be directed on a fire.


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4) Write balanced equations to describe the action of this particular fire extinguisher. What is the role of the surfactant additive?

5) Calculate the minimum mass percent concentration of sulfuric acid needed for the reaction with the gas-forming substance in the main chamber.

6) Estimate the volume of foam (in dm3 at 20°C and 1 atm) that would be produced by a fire extinguisher of this particular model.

Problem 11-4 The following method is used for quantitative determination of hydrogen sulfide in air. The analyzed sample is bubbled through an excess of aqueous cadmium acetate. The result-ing precipitate is treated with an excess of aqueous iodine whose concentration is known. This reaction proceeds quantitatively. The unreacted iodine is then titrated with a standard-ized solution of sodium thiosulfate using starch as an indicator. In one particular experiment, a 2.68 dm3 sample of air (measured at 24.0°C and 98.0 kPa) was bubbled through an excess of aqueous cadmium acetate. The resulting precipi-tate was treated with 20.0 cm3 of a 0.08226 M solution of iodine. Titration of this mixture required 15.80 cm3 of a 0.1085 M aqueous solution of sodium thiosulfate.

1) Write balanced ionic equations to represent the reactions that occur in this analy-sis.

2) What changes of color can be observed in the course of this analysis? How would you explain each of them?

3) Calculate the volume of H2S (at S.T.P.) in the air sample. 4) Calculate the mass percent of H2S in the analyzed air. 5) Express the concentration of H2S in the analyzed air in parts per million (ppm).

Problem 11-5 Many properties of atoms and molecules can be rationalized by introducing the concept of an effective nuclear charge felt by an electron in a particular orbital. This charge, denoted by Zeff, can be determined by a set of rules developed by John Slater.

According to Slater's rules, the electrons are divided into groups that keep s- and p- orbitals with the same principal quantum number together, but otherwise are made up of individual subshells: [1s][2s,2p][3s,3p][3d][4s,4p][4d][4f][5s,5p][5d], and so on. Each group is assigned a different shielding constant which depends upon the number and types of electrons in the groups preceding it. The shielding constant for each group is formed as the sum of the following contributions: (i) An amount of 0.35 from each other electron within the same group except for the [1s] group, where the other electron contributes only 0.30. (ii) If the group is of the [ns,np] type, an amount of 0.85 from each electron with a principal quantum number (n–1) and an amount of 1.00 for each electron with principal quantum numbers (n–2) and smaller. (iii) If the group is of the [nd] or [nf], type, an amount of 1.00 for each electron with a principal quantum number (n−1) and smaller. The effective charge is the difference between the actual nuclear charge and the sum of these shielding constants.

For example, the nuclear charge of the carbon atom is 6 and the electrons belong to one of the two groups, [1s] and [2s,2p]. The effective nuclear charges experienced by the electrons in the 1s, 2s, and 2p orbitals are as follows:

Zeff(1s) = 6 – (1 × 0.3) = 5.7 Zeff(2s) = 6 – (2 × 0.85 + 3 × 0.35) = 3.25 Zeff(2p) = 6 – (2 × 0.85 + 3 × 0.35) = 3.25

1) Write the electron configurations of the oxygen, sodium, and titanium atoms. 2) Calculate the effective nuclear charge experienced by electrons occupying the

highest and next-to-highest energy levels in each of these atoms. The relative ability of an atom to attract valence electrons is called electronegativity. Quan-


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titatively, an atom's electronegativity χ can be estimated using a formula proposed by All-red and Rochow: χ = 0.359Z*/r2

cov + 0.744, where Z* is the effective nuclear charge experi-

enced by valence electrons and rcov is the covalent radius of the atom expressed in Ång-strøms (Å). (1 Å = 10–10 m).

3) Calculate the Allred–Rochow electronegativity of the atoms H, O, F, and Na.. Atom H O F Na rcov, pm 42 74 72 167

4) Estimate the polarity of the O–H bond in a water molecule relative to NaF by as-suming that (a) the polarity of a bond is proportional to the difference in the elec-tronegativities of the bonded atoms; (b) the valence electron of the sodium atom in the NaF molecule is completely transferred to the fluorine.

5) Estimate the dipole moment of the water molecule relative to the dipole moment of NaF. The dipole moment of a molecule is the vector sum of the dipole mo-ments of the individual bonds. The dipole moment μ of a bond is μbond = lbondq, where the charge q is equal to the relative polarity of the bond. The H–O–H angle in the water molecule is 105°. The relative dipole moment of NaF is 7.2 arbitrary units.


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SOLUTIONS GRADE 9

Problem 9-1 1) When a pure solvent freezes, the temperature remains constant until all of the solvent has crys-

tallized. 2) As the solvent precipitates, the molality of the solution increases, so the freezing point becomes

depressed even lower. Therefore, when the solvent crystallizes, the temperature of the solution decreases.

3) Since NaCl is completely ionized in aqueous solution, the number of particles is doubled: NaCl = Na+ + Cl–.

Therefore, the molality of the solution is 21

86.1500.0

× = 0.1344 mol/kg, that is, the solution con-

tains 0.1344 mol of NaCl per 1 kg of water. M(NaCl) = 58.5 g/mol.

The mass percent of NaCl in the solution is 5.581344.01000

5.581344.0×+

× × 100% = 0.780%.

4) The enthalpy of fusion is ΔmH = RTm2M/K.

The melting point of pure water is 0оС or 273 K. M(H2O) = 18 g/mol = 0.018 kg/mol. Therefore,

ΔmH = RTm2M/K =

86.1018.0)273(314.8 2 ×× = 5996 J/mol ≈ 6 kJ/mol

5) Let us relate the mole fraction of the solute, x, to the molality of the solution, b. By definition, x = n(solute)/[n(solute) + n(solvent)]. For dilute solutions, n(solute) << n(solvent), so we can write x ≈ n(solute)/n(solvent) = n(solute) × [M(solvent)/m(solvent)] = [n(solute)/m(solvent)]M = bM, where b is the molality of the solution and M is the molar mass of the solvent. Combining equation 2 with the equation for the cryoscopic constant we have

ΔT = sm

m xH

RTΔ

2

= MK

sx = MK

bM = Kb.

6) Let the mole fraction of the impurity in the sample be x. Then the mole fraction of the impurity in the liquid phase is x/y, where y is the fraction of the fused substance. We have following sys-tem of equations:

(Tm – 181.85) = H

RT

m

m

Δ

2

28.0sx

.

(Tm – 182.25) = H

RT

m

m

Δ

2

53.0sx

, where Tm is the melting point of the pure substance.

Solution of this system of equations yields the melting point of the pure substance, Tm(pure) = 182.70 K, and the mole fraction of the impurity, x = 0.0087 or 0.87%. The melting point of the sample corresponds to y=1.00. Therefore, Tm(sample) = 182.25 + (RTm

2/ΔmH) × (x/1.00) = 182.46 K.

Problem 9-2 1) Mineral phosphorus occurs in the form of PO4

3− anions. Therefore, we can write the composi-tion of pyromorphite as Pbx(PO4)yClz. M(P) = 30.97 g/mol. M(Cl) = 35.45 g/mol.

y : z = 97.30

851.6 : 45.35

614.2 = 0.22121 : 0.07374 = 3 : 1.


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This gives Pbx(PO4)3Cl. Assuming that lead in pyromorphite is bivalent we obtain x = 5. Thus, the composition of pyromorphite is Pb5(PO4)3Cl. Let us assume that the two other miner-als have the composition Pb5X3Cl, where Х is a trivalent atom or a group of atoms.

The molar mass of mimetite (Pb5X3Cl) is 02382.0

45.35 = 1488.2 g/mol.

M(Pb) = 207.2 g/mol.

Thus, the molar mass of the group Х is 3

35.45 - 207.25 - 1488.2 × = 138.91 g/mol.

The group Х is AsO4. The formula of mimetite is Pb5(AsO4)3Cl.

The molar mass of the third mineral, Pb5Y3Cl, is 02503.0

45.35 = 1416.3 g/mol.

The molar mass of the group Y is 3

35.45 - 207.25 - 1416.3 × = 114.95 g/mol.

The group Y is VO4 ; “brown lead” is Pb5(VO4)3Cl. 2) Reaction of pyromorphite with an excess of dilute hydrochloric acid:

Pb5(PO4)3Cl + 9HCl = 5PbCl2↓ + 3H3PO4. Reaction with an excess of concentrated HCl: Pb5(PO4)3Cl + 19HCl = 5H2[PbCl4] + 3H3PO4.

3) Pb5(VO4)3Cl → 3V. M(V) = 50.94 g/mol. M(Pb5(VO4)3Cl) = 1416.3 g/mol.

The ore contains 10(0.88) = 8.8 tons or 3.1416

108.8 6× = 6.21 × 103 mol of Pb5(VO4)3Cl.

The ideal yield is 6.21 × 103 × 3 = 1.86 × 104 mol of vanadium metal. The overall practical yield is 1.86 × 104 (0.86)4 = 1.167 × 104 mol or 1.167 × 104 × 50.94 = 5.94 × 105 g = 594 kg of vanadium metal.

Problem 9-3 1) The three hydrocarbons are

CH3 CH3CH2 CH2 CH CH

ethylene acetyleneethane . Acetylene can be prepared by treating calcium carbide with water: CaC2 + 2H2O → Ca(OH)2 + C2H2. Ethylene can be obtained by treating ethanol with concentrated sulfuric acid: C2H5OH → C2H4 + H2O. Ethane can be prepared by heating a salt of propionic acid with an alkali, e.g., C2H5CO2Na + NaOH → Na2CO3 + C2H6.

2) The combustion reactions are:

C2H2 + 25

O2 → 2CO2 + H2O;

C2H4 + 3O2 → 2CO2 + 2H2O;

C2H6 + 27

O2 → 2CO2 + 3H2O.

The enthalpy change for the first reaction is

ΔrH0 = 2ΔfH0(CO2(г)) + ΔfH0(H2O(г)) – ΔfH0(C2H2(г)) – 25

ΔfH0(O2(г)) =

= 2 × (-395) + (-242) – 227 – 25

× 0 = -1259 kJ.

Similarly, for the second and the third reactions,


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ΔrH0 = 2ΔfH0(CO2(г)) + 2ΔfH0(H2O(г)) – ΔfH0(C2H4(г)) – 3ΔfH0(O2(г)) = = 2 × (-395) + 2 × (-242) – 52 – 3 × 0 = -1326 kJ.

ΔrH0 = 2ΔfH0(CO2(г)) + 3ΔfH0(H2O(г)) – ΔfH0(C2H6(г)) – 27

ΔfH0(O2(г)) =

= 2 × (-395) + 3 × (-242) – (-85) – 27

× 0 = -1431 kJ.

3) We write

CnH2n + 23n

O2 → nCO2 + nH2O + Q.

The optimal mole ratio of the reagents is CnH2n : O2 = 1 : 23n

.

4) The thermochemical equations are

C2H2 + 25

O2 → 2CO2 + H2O + 1259 kJ

C2H4 + 3O2 → 2CO2 + 2H2O + 1326 kJ

C2H6 + 27

O2 → 2CO2 + 3H2O + 1431 kJ

The amount of heat produced by combustion of 1 mol of a stoichiometric mixture is

C2H2: 251

1

+ × 1259 = 359.7 kJ

C2H4: 311+ × 1326 = 331.5 kJ

C2H6: 271

1

+ × 1431 = 318 kJ

Therefore, the ratio of the amounts of heat is acetylene : ethylene : ethane = 1.00 : 1.04 : 1.13.

5) Acetylene, because it produces the least amount of water.

Problem 9-4 1) Metal hydroxides decompose when heated:

2Me(OH)n → Me2On + nH2O↑. If the resulting oxide is not very stable, the metal can change its oxidation state. There are two possibilities to consider. Decomposition of the oxide lowers the oxidation number:

Me2On → Me2Om + 2mn −

O2↑;

Reaction of the oxide with water vapors increases the oxidation number of the metal: Me2On + (p – n)H2O → Me2Op + (p – n)H2↑. At 400 K and 110 kPa, water is in the gas phase, so the gaseous products can be either (O2 + H2O) or (H2 + H2O). М(H2) = 2.02 g/mol. М(H2O) = 18.02 g/mol. М(O2) = 32.00 g/mol. From the ideal gas equation of state

110400314.8)101020.4( 34 ××××

====−

pRT

pVmRT

nmM ρ = 12.70 g/mol,

which means that the gaseous mixture is (H2 + H2O).


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Let the mole fraction of water vapor in this mixture be x. Then we have the equation 18.02х + 2.02(1 – х) = 12.70, from which х = 0.667 х(H2O) = 0.667 х(H2) = (1 – 0.667) = 0.333 The mass fractions of the ingredients are

w(H2O) = 70.12

667.002.18 × = 0.947

w(H2) = (1 – 0.947) = 0.053 2) On the basis of the analysis of part (1) we can conclude that the hydroxide undergoes the

reaction xMe(OH)a → xMeOb + 2H2O↑ + H2↑ The oxidation number of Me in the oxide is higher than in the hydroxide. Let us write the formula of the oxide as Me2On, where n is the oxidation number of the metal in the oxide. Then

0.276 = 2)(00.16

00.16×+ MeAn

n

r

,

which gives Ar(Me) = 21.0n. There are no metals with relative atomic masses equal to 21n, where n is an integer. Therefore, the oxide must be mixed, for instance, Me3O4. Assuming this composition, we have

0.276 = 3)(400.16

400.16×+×

×MeAr

,

from which Ar(Me) = 55.9, so that Me = Fe. If B is Fe3O4, then А is Fe(OH)2. The reaction of thermal decomposition of iron(II) hydroxide is 3Fe(OH)2 → Fe3O4 + 2H2O↑ + H2↑.

3) Fe + 2HCl → FeCl2 + H2↑; FeCl2 + 2NaOH → 2NaCl + Fe(OH)2↓.

Problem 9-5 1) CaCO3 + 2HCl = CaCl2 + H2O + CO2↑;

HCl + NH3 = NH4Cl; CaCl2 + Na2C2O4 =- CaC2O4↓ + 2NaCl; Ca2C2O4 + H2SO4 = CaSO4↓ + H2C2O4; 5H2C2O4 + 2KMnO4 + 3H2SO4 = K2SO4 + 2MnSO4 + 10CO2↑ + 8H2O.

2) The titration of the 10.00 mL sample required

0.1844 × 1000

85.14 = 2.738 mmol of KMnO4.

The sample contains 2.738 × 25

= 6.846 mmol of oxalic acid, H2C2O4.

100 mL of the solution contains 6.846 × 10 = 68.46 mmol of H2C2O4. The amount of sodium oxalate added to the 25.00 mL sample is

2.000 × 100050

= 0.1000 mol = 100.0 mmol of Na2C2O4.

Of this amount, (100.0 – 68.46) = 31.54 mmol of C2O42– anions reacted with calcium ions.

Therefore, the 25.00 mL sample contained 31.54 mol of Ca2+. M(CaCO3) = 100.1 g/mol. 100 mL of the solution contains 31.54 × 4 = 126.2 mmol of Ca2+ or 126.2 mmol or 0.1262 × 100.1 = 12.63 g of CaCO3.


– 13 –

The mass percent of calcium carbonate in the limestone sample is 64.1463.12 × 100% = 86.3 %

GRADE 10

Problem10-1 1) The empirical formula of the hydrocarbon X is

N(C) : N(H) = 008.1

49.89100:01.1249.89 − = 7.45 : 10.43 = 5 : 7.

The number of hydrogen atoms in a hydrocarbon cannot be odd. Therefore, the simplest mo-lecular formula of X is C10H14. М(C10H14) = 134 g/mol From the vapor density data, the molecular mass of X does not exceed

3.101)25273(314.810 +××

====pRT

pVmRT

nmM ρ = 245 g/mol.

Therefore, Х is C10H14. 2) The two possible structures of X are

tert-butyl benzene durene . 3) Only arenes with hydrogen atoms in the α-position can be oxidized with KMnO4. Therefore,

durene can be ruled out. Compound X is tert-butyl benzene 4) The tert-butyl group directs the entering nitro group to the para-position (the ortho-position is

sterically hindered). The carboxyl group is a meta-director. Thus, the reactions are

O2N

CO2H

O2N

CO2H

NO2

CO2H

HNO3/H2O

HNO3/H2OHNO3/H2SO4

HNO3/H2SO4

5) The only isomer of X that exhibits optical activity is 2-phenyl butane:

R S .

Problem 10-2 1) The general formula of oxoacids of phosphorus is HhPpOo.

The empirical formula of Х is:


– 14 –

N(H) : N(P) : N(O) = 0.16

9.4658.4100:0.319.46:

01.158.4 −− = 4.53 : 1.51 : 3.03 = 3 : 1 : 2.

Thus, X is H3PO2 (hypophosphorous acid). Assuming that Y has the composition HhPpOo and using the data of the elemental analysis we have

1.0h + 31.0p + 16.0o = 378.0

0.31 p = 82.0p,

from which 1.0h + 16.0o = 51.0p. The only solution in integers is h = 3, p = 1, o = 3. Thus, Y is H3PO3 (phosphorous acid). The molecule of Z contains 7(1 + 0.143) = 8 atoms. Z is H3PO4 (phosphoric acid).

2) The structural formulas of the three acids are

OH P O

OH

OH

OH P O

OH

H

OH P O

H

HOH

P

OHOH

(*) Z Y X . The first dissociation constants of these acids decrease in the order Х > Y ∼ Z. (The actual pKa values are 1.2, 2.1, and 2.1, respectively.)

3) Complete neutralization of the three acids is described by the equations H3PO2 + NaOH = NaH2PO2 + H2O; H3PO3 + 2NaOH = Na2HPO3 + 2H2O; H3PO4 + 3NaOH = Na3PO4 + 3H2O. Acids X and Y react with iodine according to the equations: H3PO2 + 2I2 + 2H2O = H3PO4 + 4HI; H3PO3 + I2 + H2O = H3PO4 + 2HI. One of these reactions is fast at room temperature, but the other requires heating. Another possibility to consider is the stepwise oxidation: H3PO2 + I2 + H2O = H3PO3 + 2HI; H3PO3 + I2 + H2O = H3PO4 + 2HI. Titration of iodine with thiosulfate is based on the reaction I2 + 2Na2S2O3 = 2NaI + Na2S4O6.

4) Let the amounts of acids X, Y, and Z in the 50 mL sample be х, у, and z mmol, respectively. М(NaOH) = 40.0 g/mol. The complete neutralization requires

9.35 × 1.07 × 0.100 = 1.00 g of NaOH, or 0.40

00.1 = 0.0250 mol = 25.0 mmol

Then (x + 2y + 3z) = 25.0 The amount of iodine added to samples 1 and 2 is 200 × 0.100 = 20.0 mmol. The titration required: For sample 1: 80 × 0.100 = 8.00 mmol of sodium thiosulfate; For sample 2: 360 × 0.100 = 36.0 mmol of sodium thiosulfate The amounts of iodine reacted with the acids are:

Sample 1: (20.0 – 200.8 ) = 16.0 mmol

Sample 2: (20.0 – 2

0.36 ) = 2.0 mmol

It is known that Y reacts with iodine fast, whereas X reacts slowly and only when the reaction


– 15 –

mixture is heated. Thus, we have x + 2y + 3z = 25.0 y = 2.0 2x + y = 16.0 The solution of this system of equations is x = 7.0, y = 2.0, z = 4.67. The concentrations of the acids in the sample are

с(H3PO2) = 50

0.7 = 0.14 М; с(H3PO3) = 50

0.2 = 0.040 М; с(H3PO4) = 5067.4 = 0.093 М.

Let us now show how one can arrive at the same conclusion even without knowing relative oxi-dation rates for the acids. Suppose X is oxidized by iodine fast and that Y is oxidized slowly. This assumption leads to the system of equations x + 2y + 3z = 25.0 2x = 2.0 2x + y = 16.0 Solution of this system yields z < 0, which means that the starting assumption is incorrect. Assuming that the oxidation of X to Y is fast, while the oxidation of Y to Z is slow, we have x + 2y + 3z = 25.0 x = 2.0 2x + y = 16.0 This also gives z < 0.

5) One possibility is 2P4 + 3Ba(OH)2 + 6H2O = 3Ba(H2PO2)2 + 2PH3↑; Ba(H2PO2)2 + H2SO4 = BaSO4↓ + 2H3PO2.

Problem 10-3 1) The reactions are

3H2 + N2 = 2NH3; NH3 + HCl = NH4Cl; HCl + NaOH = NaCl + H2O; 4NH3 + 3O2 = 2N2 + 6H2O; 2H2 + O2 = 2H2O; 3H2O + P2O5 = 2H3PO4.

2) M(HCl) = 36.46 g/mol. The initial solution contains 0.0439 × 10 × 1.020 = 0.4478 g or

46.364478.0 = 12.28 mmol of HCl. The neutralization of the excess of HCl required

0.5486 × 1000

30.18 = 10.04 mmol of NaOH.

The amount of HCl reacted with ammonia is (12.28 – 10.04) = 2.24 mmol HCl. The amount of NH3 absorbed by the solution is also 2.24 mmol. M(NH3) = 17.03 g/mol. Therefore, the mass of the solution increased by 0.00224 × 17.03 = 0.0381 g and became (10 × 1.020 + 0.0381) = 10.24 g. The amount of HCl remaining in the solution is 10.04 mmol.

The mass percent of the remaining HCl is 24.101000

04.1046.36 ×× 100% = 3.57%.

3) The mass of the column filled with P2O5 increased due to absorption of water. M(H2O) = 18.02 g/mol.


– 16 –

The amount of water in one half of the reaction mixture is 02.18

08438.0 = 4.683 mmol.

Therefore, the whole mixture contains 4.683 × 2 = 9.366 of hydrogen gas.

The initial mixture contained a total of 4.22

500.0 = 22.32 mmol of N2 and H2.

The amount of N2 is (22.32 – 9.366) = 12.95 mmol. M(H2) = 2.016 g/mol M(N2) = 28.01 g/mol. The mass percent of hydrogen gas in the initial gaseous mixture is

95.1201.28366.9016.2366.9016.2

×+×× × 100% = 4.95 %.

Problem 10-4 1) 3Fe3P + 41HNO3 = 3H3PO4 + 9Fe(NO3)3 + 14NO↑ + 16H2O or

Fe3P + 23HNO3 = H3PO4 + 3Fe(NO3)3 + 14NO2↑ + 10H2O; H3PO4 + 12(NH4)2MoO4 + 21HNO3 = (NH4)3H4[P(Mo2O7)]6↓ + 21NH4NO3 + 10H2O; (NH4)3H4[P(Mo2O7)]6 + 23NH3·H2O = 12(NH4)2MoO4 + (NH4)2HPO4 + 13H2O; (NH4)2HPO4 + NH3·H2O + MgCl2 = MgNH4PO4·6H2O↓ + 2NH4Cl + H2O; 2MgNH4PO4·6H2O = Mg2P2O7 + H2O↑.

2) If the oxidizing agent is not sufficiently strong, the following reaction can occur 2Fe3P + 12H+ = 6Fe2+ + 2PH3↑ + 3H2↑. Some phosphorus can escape in the form of PH3 gas.

3) M(Mg2P2O7) = 222.6 g/mol

The amount of Mg2P2O7 is 6.222

1796.0 = 0.8068 mmol

Therefore, the initial sample contained 0.8068 × 2 = 1.614 mmol of phosphorus. M(P) = 30.97 g/mol. The mass percent of elemental phosphorus in the initial sample of cast iron is

66.14)1000/614.1(97.30 × × 100% = 0.3410%.

Problem 10-5 1) Let us write the empirical formula of Х as СaBy.

Each cubic unit cells contains 8 × 81

= 1 calcium atom and y boron atoms.

The volume of the unit cell is (415 × 10–10 cm)3 = 7.15 × 10–23 cm3. Given the density of 2.44 g/cm3, the mass of the unit cell is 2.44 × (7.15 × 10–23) = 1.744×10–22 g. M(Ca) = 40.08 g/mol.

The mass of one calcium atom is 231002.608.40

× = 6.658 × 10–23 g.

The mass of y boron atoms is (1.744 × 10–22 – 6.658 × 10–23) = 1.078 × 10–22 g. M(B) = 10.81 g/mol.

The mass of one boron atom is 231002.681.10

× = 1.796 × 10–23 h.

Therefore, y = 23

22

10796.110078.1

−

−

×× = 6.

Х is CaB6, calcium hexaboride. 2) Calcium metaborate, Ca(BO2)2, and calcium tetraborate, CaB4O7. These compounds have a


– 17 –

lower density than copper, so they float and can be easily removed. 3) CaO + 3B2O3 + 10Mg = CaB6 + 10MgO (~1100 оС);

Ca(OH)2 + 7B = CaB6 + BO↑ + H2O↑ (~1000 оС); CaCl2 + 6NaBH4 = CaB6 + 2NaCl +12H2↑ (~500 оС).

4) The six boron atoms at the center of the unit cell are arranged in the form of an octahedron.

GRADE 11

Problem 11-1 1) The empirical formula of the hydrocarbon A is

N(C) : N(H) = 01.1

3.92100:0.123.92 − = 7.69 : 7.63 = 1 : 1.

Therefore, the molecular formula of A is (CH)2n. М[(CH)2n] = 26n g/mol. Since M does not exceed 4 × 22.4 = 90 g/mol, we conclude that n ≤ 3. Because A reacts with a nitrating mixture, it is probably an arene. This implies that n is at least 3. Therefore, А = C6H6 (benzene), B is nitrobenzene, and C is aniline hydrochloride:

NO2 NH3 NH2+Cl-

A - benzene C - aniline hydrochlorideВ - nitrobenzene

HNO3

H2SO4

Fe/HCl 1) NaOH

Assuming that D contains s atoms of sulfur, we write

М(D) = s1405.0

07.32 = 228.3s (g/mol).

For each sulfur atom in D, there are 01.14

1227.08.228 × = 2 nitrogen atoms. Since D is produced by

the reaction of aniline and hydrogen sulfide, the other elements comprising D can be only car-bon and hydrogen. We have: M(D) = 228.3 = 1 × 32.07 + 2 × 14.01 + N(C) × 12.01 + N(H) × 1.01. The only solution in integers is C13H12N2S. We have aC6H7N + bCS2 → dC13H12N2S + ? Assuming that the second product is H2S, the equation becomes balanced if a = 2, b = d = 1. This reaction is similar to the reaction of ammonia with carbon dioxide. This suggests that D is N,N'-diphenylthiourea. The next step is C13H12N2S + HCl → C6H8NCl + E. Compound E must be phenyl isothiocyanate. The latter reacts with ammonia to give the final product, phenylthiourea (the mass percent of carbon is 7(12.01)/152.2=0.5524):

NH

NH2

S

NH3

+ KCl

+Cl-

+ KSCN t0

. 2) X is phenylthiourea. 3) М(A) = 78.1 g/mol.

М(X) = 152 g/mol. Synthesis of 1 mol of phenylthiourea requires 2 mol of benzene. Thus, the ideal yield of X is

15221

1.7800.1

×× = 0.973 kg.

4) The reaction is of the type:


– 18 –

C6H8NCl + Y (a carbon-containing potassium salt) → C7H8N2S + … It is obvious that the salt Y contains sulfur in addition to K and C, and that the reaction products contain potassium and chlorine. If each formula unit of Y contains x carbon atoms, then the molecular mass of this compound is

М(Y) = x1236.0

01.12 = 97.1x (g/mol)

The only possibility is that Y is potassium thiocyanate, KSCN. The reaction is:

NH2NH

NH

S

N C S

NH

NH2

SX

CS2, t0 HCl, t0

NH3

- C- H2S

. This process is similar to the first synthesis of an organic substance from inorganic reagents: the preparation of urea from silver cyanate and ammonium chloride by Friedrich Wöhler in 1828.

Problem 11-2 1) The binary compound А contains hydrogen and is a strong reducing agent. Therefore, it is some

hydride. Reaction of A with water produces hydrogen gas (compound B). Let the mass of one equivalent of the unknown element X be Me(X). From the condition that 1.0/(1.0 + Me(X)) > 0.10 we obtain the inequality Me(X) < 9.0. Since the hydride is solid, the only possibilities are the metals lithium and beryllium. Since C is the only product of the reaction of A and B, we can write XHx + aCO2 = XHx⋅aCO2,

from which 616.00.44)(0.1

0.162=

×++××

aXMxa

e

, so Me(X) = (7.94а – 1.0х) g/mol.

The value of x must be a small integer, while a must be a rational number. The physically ac-ceptable solution is а = х = 1. Then Me(X) = 6.94 g/mol, which means that X is lithium. А is LiH (lithium hydride). В is LiH⋅CO2 (or, more correctly, HCOOLi), lithium formiate. Reaction of a formiate with dilute sulfuric acid produces formic acid (compound D). Concen-trated sulfuric acid dehydrates HCOOH and yields carbon monoxide (compound E). D is HCOOH (formic acid); E is CO (carbon monoxide).

2) The reactions are as follows: LiH + H2O → LiOH + H2↑; LiH + CO2 → HCOOLi; 2HCOOLi + H2SO4 → 2HCOOH + Li2SO4; HCOOLi + H2SO4 → CO↑ + H2O + LiHSO4.

3) Lithium hydride is an ionic compound, so its formula may be written as Li+H–. Reaction of LiH with aluminum chloride is: AlCl3 + 4LiH → Li[AlH4] + 3LiCl.


– 19 –

Lithium aluminum hydride, Li[AlH4], is widely used as a reducing agent in chemical synthesis. 4) The relative atomic masses of the elements that constitute compound A are

Ar(H) = 1.008. Ar(Li) = 6.94. The isotopes in question are 1H, 2H, 6Li, 7Li. Let the natural abundance of 1H be х. Then we have 1х + 2(1 – х) = 1.008, which gives х = 0.992. Let the natural abundance of 6Li be y. Then we have the equation 6у + 7(1 – у) = 6.94, whose solution is у = 0.06. Assuming that the probability of finding a particular isotope is proportional to its abundance, the relative abundances of the four isotopomers aLibH are 6Li1H: 0.06⋅0.992 = 0.0595 6Li2H: 0.06⋅0.008 = 0.00048 7Li1H: 0.94⋅0.992 = 0.932 7Li2H: 0.94⋅0.008 = 0.0075 The relative abundances of these four isotopomers are 0.0595 : 0.00048 : 0.932 : 0.0075 ≈ 124 : 1: 1942 : 16.

Problem 11-3 1) The calorific value of a compound is the energy released as heat when a unit mass of the com-

pound undergoes complete combustion with oxygen under standard conditions. 2) We assume that chemical bonds of the same type make equal contributions to the calorific

value of a compound. The repeating unit of polyethylene (-СН2–СН2-) contains two C–C bond and 4 C–H bonds. The repeating unit of natural rubber, -CH2–C(CH3)=CH–CH2-, contains one С=С bond, four single С–С bonds and eight С–Н bonds. The repeating unit of poly-1,3-butadiene, -CH2–CH=CH–CH2-, contains one С=С bond, three single С–С bonds, and six С–Н bonds. M(C2H4) = 28.06 g/mol M(C5H8) = 68.12 g/mol M(C4H6) = 54.09 g/mol The heat of combustion per 1 mol of repeating units is

for polyethylene: 1000

06.2814.47 × = 1.323 MJ per 1 mol of repeating units

for natural rubber: 1000

12.6873.44 × = 3.047 MJ per 1 mol of repeating units

Let us denote the number of С–С bonds by s, the number of С=С bonds by d, and the number of С–Н bonds by h. Then we can write for polyethylene and for polyisoprene 1.323 = 2s + 4h 3.047 = 4s + d + 8h. Then the heat of combustion of polybutadiene is

3s + d + 6h = 2323.1047.3 − = 2.386 MJ per 1 mol of repeating units.

Therefore, the calorific value of polybutadiene is 09.54

1000386.2 × = 44.10 MJ/kg.

There is a different way of solving this problem. Consider the combustion reactions for the re-peating units of the three polymers in question: (C2H4) + 3O2 → 2CO2 + 2H2O (C4H6) + 5.5O2 → 4CO2 + 3H2O (C5H8) + 7O2 → 5CO2 + 4H2O The first reaction involves the rupture of 2 С–С bonds, 4 С–Н bonds, and 3 О–О bonds, and the formation of 4 С=О bonds and 4 О–Н bonds. Therefore, the heat of combustion for this reac-


– 20 –

tion can be written as Q1 = 4EС=О + 4EО-Н – 2EС-C – 4EС-H – 3EO=О. Similarly, for the two other reactions we can write Q2 = 8EС=О + 6EО-Н – 3EС-C – EС=C – 6EС-H – 5,5EO=О; Q3 = 10EС=О + 8EО-Н – 4EС-C – EС=C – 8EС-H – 7EO=О.

A moment's thought shows that 21

32QQQ −= .

3) The repeating unit of cellulose is C6H10O5, for polymethyl methacrylate it is C5H8O2, for natural rubber (polyisoprene) it is C5H8, and for polystyrene it is (C8H8). The composition of all these units can be written as CxHyOz. The combustion reaction is

CxHyOz + 22

2 zyx −+O2 → xCO2 + 2

yH2O.

The calorific value of CxHyOz is

zyx

OHCQOHQyCOxQQ

zyxmf

mf

mf

1612

)()(2

)( 22

++

−+= ,

where Qfm is the heat of formation of a substance. As a rule, heats of formation of organic com-

pounds are relatively small. Also, the higher the oxygen content, the higher the heat of forma-tion. This allows us to conclude that the calorific value of a compound should decrease as the fraction of oxygen atoms in the repeating unit increases. For hydrocarbon-based polymers, the calorific value increases with increasing hydrogen content (this is because the molar heats of combustion of atomic carbon and hydrogen are similar, but carbon is an order of magnitude heavier than hydrogen). Thus, the order of increasing calorific value is: cellulose, polymethyl methacrylate, polystyrene, natural rubber. The reference values are, respectively (in MJ/kg): 13.4, 27.7, 39.0, and 44.7.

4) H2SO4 + 2NaHCO3 = Na2SO4 + 2H2O + 2CO2↑. The surfactant is added to increase the amount of soapy foam.

5) М(H2SO4) = 98 g/mol М(NaHCO3) = 84 g/mol.

The main chamber contains 8 × 0.05 = 0.4 kg = 400 g or 84400

= 4.76 mol of NaHCO3.

The stoichiometric quantity of H2SO4 is 276.498× = 233 g

Therefore, the minimum mass percent of sulfuric acid is 400233

× 100% = 58%.

6) The volume of foam is approximately equal to the volume of carbon dioxide produced by the reaction of NaHCO3 and H2SO4. Since n(CO2) = n(NaHCO3), we have

3.101

)27320(314.876.4 +××==

pnRTV = 114 dm3.

Problem 11-4 1) The reactions are:

H2S + Cd2+ + 2CH3COO– = CdS↓ + 2CH3COOH; CdS + I2 = Cd2+ + 2I– + S↓; I2 + 2S2O3

2– = 2I– + S4O62–.

2) A yellow precipitate of CdS is produced; milky-yellow colloidal sulfur is formed; the color of molecular iodine disappears during the titration.


– 21 –

3) The amount of added iodine is 0.08226 ×1000

0.20 = 1.645 mmol.

The titration of the excess iodine required 0.1085 × 1000

80.15 = 1.714 mmol of Na2S2O3,

which corresponds to 2714.1 = 0.8570 mmol of I2.

The amount of I2 reacted with CdS is (1.645 – 0.8570) = 0.7880 mmol. Therefore, the initial air sample contained 0.7880 mmol of H2S.

This amount of H2S occupies the volume 22.4 ×10007880.0 = 17.65 dm3.

4) M(H2S) = 34 g/mol.

The mass of hydrogen sulfide is 34 ×10007880.0 = 2.68 × 10–2 g.

The amount of air in the sample is

)24273(314.868.20.98+×

× = 0.1064 mol.

The average molar mass of air is M(air) = 29 g/mol. Therefore, the mass of the air sample is 0.1064 × 29 = 3.08 g.

The mass percent of H2S in the sample is 0268.008.3

0268.0+

× 100% = 0.863%.

5) 86.3 × 103 ррm by mass and 7.4 × 103 ppm by mole.

Problem 11-5 1) O: 1s22s22p4;

Na: 1s22s22p63s1; Ti: 1s22s22p63s23p63d24s2.

2) Oxygen atom: Zeff(2s,2p) = 8 – (5 × 0.35 + 2 × 0.85) = 4.55 Zeff(1s) = 8 – (1 × 0.30) = 7.70 Sodium atom: Zeff(3s) = 11 – (8 × 0.85 + 2 × 1.00) = 2.20 Zeff(2s,2p) = 11 – (7 × 0.35 + 2 × 0.85) = 6.85 Titanium atom: Zeff(4s) = 22 – (1 × 0.35 + 10 × 0.85 + 8 × 1.00 + 2 × 1.00) = 3.15 Zeff(3d) = 22 – (1 × 0.35 + 8 × 1.00 + 8 × 1.00 + 2 × 1.00) = 3.65 Zeff(3s,3p) = 22 – (7 × 0.35 + 8 × 0.85 + 2 × 1.00) = 10.75

3) The Allred–Rochow electronegativities are

H: 0.359 × 2

10

12

101042

)30.01(1

⎟⎟⎠

⎞⎜⎜⎝

⎛ ×

×−

−

− + 0.744 = 2.17

O: 0.359 × 2

10

12

101074

)85.0235.06(8

⎟⎟⎠

⎞⎜⎜⎝

⎛ ×

×+×−

−

− + 0.744 = 3.50

F: 0.359 × 2

10

12

101072

)85.0235.07(9

⎟⎟⎠

⎞⎜⎜⎝

⎛ ×

×+×−

−

− + 0.744 = 4.10


– 22 –

Na: 0.359 × 2

10

12

1010167

)00.1285.0835.01(11

⎟⎟⎠

⎞⎜⎜⎝

⎛ ×

×+×+×−

−

− + 0.744 = 0.98

4) We assume that the polarity of the chemical bond in NaF is 100 %. The difference in the elec-tronegativities of the F and Na atoms is (4.10 – 0.98) = 3.12. The difference in the electronegativities of the O and H atoms is (3.50 – 2.17) = 1.33.

The polarity of the OH bond relative to the HF bond is, therefore, 12.333.1 × 100% = 42.6 %.

5) The bond length in NaF is (167 + 72) = 239 pm. Let us assume that the charge flow in the NaF

molecule is 1.00. Therefore, our unit of the dipole moment is 2.7

239 e·pm.

The O–H bond in the water molecule has a length of (42 + 74) = 116 pm. From part (4), the charge flow for the O–H bond is 0.426. The dipole moment of the O–H bond is (116 pm)(0.426) = 49.4 e·pm, which is equivalent to 49.7/(239/7.2) = 1.49 arbitrary units.

According to the figure shown below, the magnitude of the dipole mo-ment of H2O is equal to BD, the magnitude of the vector sum of the di-pole moments of the two OH bonds. Since the angle DBC is 52.5°, we have BD = 2(BC)cos 52.5° = 2(1.49)(0.609) = 1.81 arbitrary units.

Translated by V.Staroverov

A

B

C

D

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